Find the maxima, minima, and saddle points of the function f(x,y) = x4 + y4 + 4xy.

The given function is $f(x,y) = x^4 + y^4 + 4xy$.
The partial derivative with respect to $x$ is:

$$f_x = \frac{\partial}{\partial x}(x^4 + y^4 + 4xy) = 4x^3 + 4y$$

The partial derivative with respect to $y$ is:

$$f_y = \frac{\partial}{\partial y}(x^4 + y^4 + 4xy) = 4y^3 + 4x$$

Set $f_x = 0$ and $f_y = 0$:

1. $4x^3 + 4y = 0 \implies x^3 + y = 0 \implies y = -x^3$
2. $4y^3 + 4x = 0 \implies y^3 + x = 0$

Substitute $y = -x^3$ from equation (1) into equation (2):

$$(-x^3)^3 + x = 0$$ $$-x^9 + x = 0$$ $$x(1 - x^8) = 0$$

This equation yields three possible values for $x$:

• $x = 0$
• $1 - x^8 = 0 \implies x^8 = 1 \implies x = \pm 1$

Now, find the corresponding $y$ values using $y = -x^3$:

• If $x = 0$, then $y = -(0)^3 = 0$. Critical point: $(0,0)$.
• If $x = 1$, then $y = -(1)^3 = -1$. Critical point: $(1,-1)$.
• If $x = -1$, then $y = -(-1)^3 = 1$. Critical point: $(-1,1)$.

The critical points are $(0,0)$, $(1,-1)$, and $(-1,1)$.

$$f_{xx} = \frac{\partial}{\partial x}(4x^3 + 4y) = 12x^2$$ $$f_{yy} = \frac{\partial}{\partial y}(4y^3 + 4x) = 12y^2$$ $$f_{xy} = \frac{\partial}{\partial y}(4x^3 + 4y) = 4$$

• For the critical point $(0,0)$:

  $$D(0,0) = 144(0)^2(0)^2 - 16 = -16$$

  Since $D(0,0) < 0$, the point $(0,0)$ is a saddle point.

• For the critical point $(1,-1)$:

  $$D(1,-1) = 144(1)^2(-1)^2 - 16 = 144 - 16 = 128$$

  Since $D(1,-1) > 0$, we check $f_{xx}$:

  $$f_{xx}(1,-1) = 12(1)^2 = 12$$

  Since $f_{xx}(1,-1) > 0$, the point $(1,-1)$ is a local minimum. The function value at this point is:

  $$f(1,-1) = (1)^4 + (-1)^4 + 4(1)(-1) = 1 + 1 - 4 = -2$$

• For the critical point $(-1,1)$:

  $$D(-1,1) = 144(-1)^2(1)^2 - 16 = 144 - 16 = 128$$

  Since $D(-1,1) > 0$, we check $f_{xx}$:

  $$f_{xx}(-1,1) = 12(-1)^2 = 12$$

  Since $f_{xx}(-1,1) > 0$, the point $(-1,1)$ is a local minimum. The function value at this point is:

  $$f(-1,1) = (-1)^4 + (1)^4 + 4(-1)(1) = 1 + 1 - 4 = -2$$

The final answer is:

• Saddle point at $(0,0)$
• Local minimum at $(1,-1)$ with value $-2$
• Local minimum at $(-1,1)$ with value $-2$